How to simplify in boolean algebra

I have some homework I can't seem to figure out. The assignment causing problems is devided into two parts; The first is to determine the inverse formula for a given formula (so the S = F'). The second part asks me to simplify the answer to the first question. I've looked trough all materials I have, but I can't figure out what happens here:

(I've used the ' symbol as NOT)

The first formula is as follows:

(a·c·(b + d) + b·c'·d + a'·b)


Which, when i invert it without simplifying, results into:

(a' + c' + b'·d')·(b' + c + d')·(a + b')


So far, so good. However, when I look at the given solution for the second part of the assignment, I can't seem to figure out the logic behind the given steps.

  (a' + c' + b'·d')·(b' + c + d')·(a + b')
= (a'·b' + a·c' + b'·c' + a·b'·d' + b'·d')·(b' + c + d') =
= (a'·b' + a·c' + b'·c' + b'·d')·(b' + c + d')
= a'·b' + a·b'·c' + b'·c' + b'·d' + a'·b'·c  + b'·c·d' + a'·b'·d' + a·c'·d' + b'·c'·d'+ b'·d' =
= a'·b' + b'·c' + b'·d' + a·c'·d


Can anyone point me in the direction of what steps are taken here? Thanks in advance!

Let me try this-- I have added two additional step in bold to make the steps clear.

$(a' + c' + b'·d')·(b' + c + d')·(a + b')$

$=\mathbf{(a' + c' + b'·d').(a + b')·(b' + c + d')}$ (by coomutativity of $.$)

$= (a'·b' + a·c' + b'·c' + \underline{a·b'·d' + b'·d'})·(b' + c + d')$ (multiplying and rearranging the first two factors; commutativity and associativity of $.$ and $+$, just like addition and multiplication of real number, being used )

$= (a'·b' + a·c' + b'·c' + \underline{b'·d'})·(b' + c + d')$ (using $\mathbf{x+x.y}=x:(1+y)=x.1\mathbf{=x}$ in the underlined)

= a'·b' + a·b'·c' + b'·c' + b'·d' + a'·b'·c + b'·c·d' + a'·b'·d' + a·c'·d' + b'·c'·d'+ b'·d' (multiplying and rearranging just like one step above; also $x.x=x$ has been used in all the 2-product terms)

$=\mathbf{a'.b'+\underline{b'.c'.a+b'.c'}+\underline{b'.d'+b'.d'.(a'+c+c')}+a.c'.d}$ (note that $b'.d'+b'.d'=b'.d'$ as $x+x=x\quad\forall x$; also the rearrangement is for the next step)

$= a'·b' + b'·c' + b'·d' + a·c'·d$, using the rule $x+x.y=x$ with $x=b'.c', y=a$ in the first underline and $x=b'.d', y=(a'+c+c')$ in the second underline.

• Sorry in the third line I missed the term $a′.a$ in the first factor. Please note that $a′.a=0$ and $0+a=a\quad\forall a$, so this term will have no contribution in later steps :) – Tapu Nov 1 '11 at 11:12
• Still stared at it for about an hour, but i see what happens now. The reordering makes the samples real hard to understand, mainly because is isn't stated that reordering has taken place. Thanks for explaining! – Lg102 Nov 1 '11 at 11:59
• In such cases, you are welcome to ask what is the difficulty in which step. Just for an alternate easy way to Boolean algebra, may I know if you have read basics operations of set theory? The rules are exactly the same in both cases: $(U,\phi, \cup, \cap,')$ in set theory is exactly same as $(1,0,+,.)$ in Boolean algebra. So, if have read one, the other is obviously done (this is true for basic operations though). – Tapu Nov 1 '11 at 12:26